Y = e^tanx - 2
To find at which point it crosses x axis we state that y= 0
e^tanx - 2 = 0
e^tanx = 2
tanx = ln 2
tanx = 0.69314
x = 0.6061
to find slope at that point first we need to find first derivative of funtion y.
y' = (e^tanx)*1/cos^2(x)
now we express x = 0.6061 in y' and we get:
y' = k = 2,9599
5.5x+9
Trying to make my answer 20 characters
Answer:
For x^2 + 5x - 14 = 0
x1= -7, x2 = 2
For x^2 + 5x + 6 = 0
x1 = -3, x2 = -2
Step-by-step explanation:
Pls see attached....
When you write it as an inequality you get x + 4 > 13